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Description
The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.
For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:
a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)
You are asked to write a program to find the minimum inversion number out of the above sequences.Input
The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.Output
For each case, output the minimum inversion number on a single line.Sample Input
10 1 3 6 9 0 8 5 7 4 2Sample Output
16
题意：就是给出一串数，当依次在将第一个数变为最后一个数的过程中，要你求它的最小逆序数。
题解：采用线段树维护数据。边输入，边记录每个数字之前比他小的数据。
#include <iostream> #include <cstdio> using namespace std; int sum[24000]; int min(int a, int b) { if (a > b)return b; return a; } void pushUp(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; } void build(int l, int r, int rt) { sum[rt] = 0; if (l == r) return; int m = (l + r) >> 1; build(l, m, rt << 1); build(m + 1, r, rt << 1 | 1); } void update(int p, int l, int r, int rt) { if (l == r) { sum[rt]++; return; } int m = (l + r) >> 1; if (p <= m) update(p, l, m, rt << 1); else update(p, m + 1, r, rt << 1 | 1); pushUp(rt); } int query(int L, int R, int l, int r, int rt) { if (l >= L && r <= R) return sum[rt]; int m = (l + r) >> 1, ans = 0; if (m >= L) ans += query(L, R, l, m, rt << 1); if (m < R) ans += query(L, R, m + 1, r, rt << 1 | 1); return ans; } int main() { int sum,n, s[6000]; while (scanf("%d", &n) != EOF) { build(0, n - 1, 1); sum = 0; for (int i = 0; i < n; i++) { scanf("%d", &s[i]); sum += query(s[i], n - 1, 0, n - 1, 1); update(s[i], 0, n - 1, 1); } int ans = sum; for (int i = 0; i < n; i++) { sum = sum + n - 2 * s[i] - 1; ans = min(ans, sum); } printf("%d\n", ans); } }
HDU 1394 <Minimum Inversion Number> <逆序数><线段树>
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原文地址：http://www.cnblogs.com/914295860-jry/p/5822732.html
